Minimum 2-Tuple Dominating Set of an Interval...

Hindawi Publishing CorporationInternational Journal of CombinatoricsVolume 2011, Article ID 389369, 14 pagesdoi:10.1155/2011/389369

Research ArticleMinimum 2-Tuple Dominating Set ofan Interval Graph

Tarasankar Pramanik,1 Sukumar Mondal,2and Madhumangal Pal1

1 Department of Applied Mathematics with Oceanology and Computer Programming,Vidyasagar University, Midnapore 721102, India

2 Department of Mathematics, Raja N. L. Khan Women’s College, Vidyasagar University,Midnapore 721102, India

Correspondence should be addressed to Madhumangal Pal, [email protected]

Received 8 September 2011; Revised 4 November 2011; Accepted 17 November 2011

Academic Editor: Johannes Hattingh

Copyright q 2011 Tarasankar Pramanik et al. This is an open access article distributed underthe Creative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

The k-tuple domination problem, for a fixed positive integer k, is to find a minimum size vertexsubset such that every vertex in the graph is dominated by at least k vertices in this set. The casewhen k = 2 is called 2-tuple domination problem or double domination problem. In this paper,the 2-tuple domination problem is studied on interval graphs from an algorithmic point of view,which takes O(n2) time, n is the total number of vertices of the interval graph.

1. Introduction

An undirected graph G = (V, E) is an interval graph if the vertex set V can be put into one-to-one correspondence with a set of intervals I on the real line R such that two vertices areadjacent in G if and only if their corresponding intervals have nonempty intersection. The setI is called an interval representation of G and G is referred to as the intersection graph of I [1].Let I = {i1, i2, . . . , in}, where ic = [ac, bc] for 1 ≤ c ≤ n, be the interval representation of thegraph G, ac is the left endpoint, and bc is the right end point of the interval ic. Without anyloss of generality, assume the following:

(a) an interval contains both its endpoints and that no two intervals share a commonendpoint [1],

(b) intervals and vertices of an interval graph are one and the same thing,

(c) the graph G is connected, and the list of sorted endpoints is given

(d) the intervals in I are indexed by increasing right endpoints, that is, b1 < b2 < · · · <bn.

2 International Journal of Combinatorics

In a graph G, a vertex is said to dominate itself and all of its neighbors. A dominating setofG = (V, E) is a subsetD of V such that every vertex in V is dominated by at least one vertexin D. The domination number γ(G) is the minimum size of a dominating set of G. For a fixedpositive integer k, a k-tuple dominating set of G = (V, E) is a subset D of V such that everyvertex in V is dominated by at least k vertices of D. As introduced by Harary and Haynes[2], a k-tuple dominating set D is a set D ⊆ V for which |N[v] ∩D| ≥ k for every v ∈ V , whereN[v] = {v} ∪ {u ∈ V : (u, v) ∈ E} is the closed neighborhood of the vertex v. Note that we musthave the minimum degree greater than or equal to k − 1 for a k-tuple dominating set to exist.The k-tuple domination number γ×k(G) is the minimum cardinality of k-tuple dominating setof G. When k = 2, this is called double domination [3].

A 2-tuple dominating set D is said to be minimal if there does not exist any D′ ⊂ Dsuch thatD′ is a 2-tuple dominating set of G. A 2-tuple dominating setD, denoted by γ×2(G),is said to be minimum, if it is minimal as well as it gives 2-tuple domination number.

In graph theory, a connected component of an undirected graph is a subgraph inwhichany two vertices are connected to each other by paths. For a graph G, if the subgraph G itselfis a connected component, then the graph G is called connected, else the graph G is calleddisconnected and each connected component subgraph is called its component. Removal of avertex v from a graph G means the removal of vertex v and edges incident to v. A cut vertexof a connected graph G is a vertex of G whose removal renders G disconnected. Pal et al. [4]described an algorithm for computing cut vertices and blocks on interval graphs.

A graph G is vertex domination-critical if for any vertex v of G, the domination numberof G − v is less than the domination number of G. If such a graph G has domination numberγ , it is called γ-critical. Brigham et al. [5] studied γ-critical graphs and posed the followingquestions.

(1) If G is a γ-critical graph, is |V | ≥ (δ + 1)(γ − 1) + 1?

(2) If a γ-critical graph G has (Δ + 1)(γ − 1) + 1 vertices, is G regular?

(3) Does i = γ for all γ-critical graphs?

(4) Let d be the diameter of the γ-critical graph G. Does d ≤ 2(γ − 1) always hold?

Later in this paper, it has been proved that for some vertex (or cut vertex) v of G, G−vand G have the same domination number γ×2(G).

1.1. Survey of Related Works

Various works have been found on interval graphs. Interval graphs are useful in modelingresource allocation problems in operations research. A. Pal andM. Pal [6] have studied aboutinterval graphs. So many algorithms and results of various parameters on interval graphshave been found in [4, 7–12]. The domination is one of the parameters in graphs which hasa great importance in modern circuit designing systems. Chang et al. [13] have extensivelystudied about domination in graphs. Also domination and its variations can be found in[14–17]. Another type of dominating set has been widely studied in [18] which is a totaldominating set. Henning has worked on graphs with large total domination number in [19]. Fora domination number, Sumner and Blitch [20] studied graphs where the addition of anyedge changed the domination number. They called graphs with this property domination edgecritical. Brigham et al. [5] and Fulman et al. [21] have worked on vertex domination-criticalgraphs. Wojcicka [22] have found some results onHamiltonian properties of domination-critical

International Journal of Combinatorics 3

graphs. The total domination edge critical graphs, that is, graphs where the addition of anyedge decreased the total domination number were studied by Haynes et al. in [23–26].Among the variations of domination, the k-tuple domination was introduced in [3]. Thecase when k = 2 was called double domination in [3], where exact values of the doubledomination numbers for some special graphs are obtained. In the same paper, various boundsof double and k-tuple domination numbers are available in terms of the other parameters.

2. Interval Graph and Some of Its Properties

Let G = (V, E), V = {1, 2, . . . , n}, |V | = n, |E| = m be a connected interval graph in which thevertices are given in the sorted order of the right endpoints of the interval representation ofthe graph. Intervals are labeled according to increasing order of their endpoints. This labelingis referred to as IG ordering. Let (u, v) or (v, u) denote the existence of an adjacency relationbetween two vertices u, v. It is assumed that (u, u) is always true, that is, (u, u) ∈ E. If [au, bu]and [av, bv] are two end points of the vertices u and v, respectively, then u, v are adjacent ifat least one of the following conditions hold:

(i) av < au < bv,

(ii) av < bu < bv,

(iii) au < av < bu,

(iv) au < bv < bu.

The following lemma is true for a given interval graph, G = (V, E).

Lemma 2.1 (see [27]). If the vertices u, v,w ∈ V are such that u < v < w in the IG ordering and uis adjacent to w, then v is also adjacent to w.

For each vertex v ∈ V , let H(v) represent the highest numbered adjacent vertex of v.If no adjacent vertex of v exists with higher IG number than v, thenH(v) is assumed to be v.In other words, H(v) = max{u : (v, u) ∈ E, u ≥ v}.

Throughout this paper, we use the notation D for 2-tuple dominating set. For thepurpose to find D of the interval graph G = (V, E), we consider a function f : V (G) → {0, 1}which is defined by f(v) = 1 if v ∈ D, otherwise, f(v) = 0. We define the function f so thatfor S ⊆ V (G), f(S) =

∑v∈S f(v). The weight of the function f isw(f) = f(V (G)). Also,wi(f)

is defined as wi(f) = f(N[i]) =∑

v∈N[i] f(v), for all i = 1, 2, 3, . . . , n.

3. Algorithm for 2-Tuple Domination

In a connected interval graph, the vertices are ordered by IG ordering. First of all, we treatnone of a vertex of V (G) as a member of dominating set D. Then, insert vertices one by oneby testing their consistency. If a vertex v is dominated by at least two vertices then leave it,otherwise, take the highest numbered adjacent vertex (vertices) from N[v] as member(s) ofdominating set D.


1

2

3 5

4 6

7

8

9

10

Figure 1: An interval graph G = (V, E).

Table 1: Computation of all pth numbered adjacent vertices.

Mi(v)v

1 2 3 4 5 6 7 8 9 10M0(v) 5 5 5 8 8 8 8 10 10 10M1(v) 4 4 4 5 6 7 7 8 9 9M2(v) 1 2 3 4 5 6 6 7 — 8M3(v) — — — 3 4 5 — 6 — —M4(v) — — — 2 3 — — 5 — —M5(v) — — — 1 2 — — 4 — —M6(v) — — — — 1 — — — — —

Let us associate a new term Mi(v) for a vertex v ∈ V , for all i = 0, 1, 2, . . . , k (k =|N(v)|) to each adjacent vertices of v in order to set IG ordering of intervals in the followingway:

Mi(v) = max

⎧⎨

⎩N[v] −

i−1⋃

j=0

Mi(v)

⎫⎬

⎭

with M0(v) = max{N[v]}.

(3.1)

Basically,M0(v) = H(v), the highest numbered adjacent vertex of v [28]. In connectionwith the name of H(v), we call this Mi(v) as the pth numbered adjacent vertex of v throughDefinition 3.1.

Definition 3.1 (pth numbered adjacent vertex). Let u, v ∈ V . If for some i (i =0, 1, 2, . . . , |N(v)|), |N(v)| − i = p such that u = Mi(v), then u is called the pth numberedadjacent vertex of v.

From the definition, it is easily seen that, for a vertex v, Mi(v) exists for maximumpossible i = |N(v)|, that is, degree of the vertex v. Therefore, in a graph, the maximumpossible i occurs in the degree of the graph, that is, Δ = max{deg(v) : v ∈ V }. An illustrationof the computations of all Mi(v) for the graph of Figure 1 are shown in Table 1.

Now, we describe an algorithm to find two sets of verticesD and L depending only onM0(i) and M1(i).


Input: An interval graph G = (V, E) with IG ordering vertex set V = {1, 2, 3, . . . , n}.Output: 2-tuple dominating set D and 2-tuple domination number γ×2(G)(= |D|).Step 1: Set f(j) = 0, ∀j = 1, 2, . . . , n; //Assume that no vertices are the members

of D.//Step 2: Set i = 1, D = ∅ and L = ∅;

Step 2.1: Compute wi(f) =∑

v∈N[i] f(v);Step 2.2: If wi(f) = 0 then //At least the vertex i is not adjacent to any of the

vertices of D.//Set f(M0(i)) = 1 and f(M1(i)) = 1;D = D ∪ {M0(i)} ∪ {M1(i)} and L = L ∪ {i};

else if wi(f) = 1 then //At least the vertex i is connected to one of thevertex of D.//

If f(M0(i)) = 1 thenSet f(M1(i)) = 1;D = D ∪ {M1(i)};

elseSet f(M0(i)) = 1;D = D ∪ {M0(i)};

end if;L = L ∪ {i};

elseGoto Step 2.3;

end if;Step 2.3: Calculate i = i + 1 and go to Step 2.1 and continue until i > n;

end 2DIG.

Algorithm 1: Algorithm 2DIG.

Actually, the Algorithm 2DIG (Algorithm 1) gives the set D which is the minimum 2-tuple dominating set and |D|, the 2-tuple domination number of the interval graphG = (V, E).Before going to prove this result, we first verify Algorithm 2DIG in Figure 2. Here, we denotethe set L as the set of leading vertices corresponding to the 2-tuple dominating set D.

In Algorithm 2DIG, at ith iteration, if wi(f) = 0, then i is a member of L and i is saidto be the leading vertex of order 2 corresponding to the vertices M0(i) and M1(i) of D, and ifwi(f) = 1, then i is said to be the leading vertex of order 1 corresponding to the vertexM0(i) orM1(i) of D, otherwise, i does not belong to L.

Therefore, we conclude that if l1 ∈ L, then l1 is adjacent to exactly two vertices of D.

3.1. Verification of the Algorithm

Suppose we are to find 2-tuple dominating set D and 2-tuple domination number |D| of theinterval graph G = (V, E), where V = {1, 2, . . . , 10} shown in Figure 1. First, set f(j) = 0, forall j ∈ V . In Step 2, set i = 1, D = ∅ and L = ∅, that is, initially D and L are empty. Step 2repeats for n times. Here, n = 10, number of vertices in the graph. We illustrate the iterationsin the following way.

Iteration 1. For the first iteration i = 1,N[1] = {1, 4, 5}. Calculatew1(f) = f(1)+f(4)+f(5) = 0.The first condition of if-end if is satisfied. Since w1(f) = 0, we find M0(1) = 5 and M1(1) = 4.Then, set f(5) = 1 and f(4) = 1. Also, set D = ∅ ∪ {4, 5} = {4, 5}, L = {1}, and i = i + 1 = 2.


1 2 3

5

4 6

7

8 9

10

Figure 2: An interval representation of Figure 1.

1 2 3

5

4 6

7

8 9

10

Figure 3: Finding D by Algorithm 2DIG.

Iteration 2. N[2] = {2, 4, 5}. w2(f) = f(N[2]) = 2. The vertex 2 is dominated by two vertices4 and 5 ofD. So, in this iteration,D could not be calculated. Hence,D and L remain the sameand i is being increased to 3.

Iteration 3. N[3] = {3, 4, 5}. w3(f) = f(N[3]) = 2. In this iteration, also D and L remain un-changed. The iteration number i is being increased to 4.

Iteration 4. Here, N[4] = {1, 2, 3, 4, 5, 8} and w4(f) = f(N[4]) = 2. So, D and L are the sameas the previous iteration. Set i = 5.

Iteration 5. In this iteration, N[5] = {1, 2, 3, 4, 5, 6, 8} and w5(f) = f(N[5]) = 2, and hence nochange occurs. i is being increased to 6.

Iteration 6. N[6] = {5, 6, 7, 8} and w6(f) = f(N[6]) = 1. So, domination criteria are notsatisfied here. The else-if condition of if-end if is satisfied. Now, we check either f(M0(6)) = 1or not.We see that f(M0(6)) = f(8) = 0, and hence set f(8) = 1. UpdateD byD∪{8} = {4, 5, 8}and L by L ∪ {6} = {1, 6}. i is being increased to 7.

Iteration 7. N[7] = {6, 7, 8} and w7(f) = f(N[7]) = 1. Here, also domination criteria are notsatisfied. As f(M0(7)) = f(8) = 1, set f(M1(7)) = f(7) = 1 and D = D ∪ {7} = {4, 5, 8, 7} withL = L ∪ {7} = {1, 6, 7}. i is being increased to 8.

Iteration 8. In this iteration, that is, for i = 8,N[8] = {4, 5, 6, 7, 8, 10}. w8(f) = 4. Hence, D andL remain unchanged and i is being increased to 9.

Iteration 9. At ninth iteration, i = 9. Here, N[9] = {9, 10} w9(f) = 0. Then, D = D ∪ {9, 10} ={4, 5, 8, 7, 9, 10} and L = L ∪ {9} = {1, 6, 7, 9} with f(9) = 1 and f(10) = 1. Set i = 10.

Iteration 10. For i = 10, N[10] = {8, 9, 10} w10(f) = 3. Hence, D and L remain unchanged. Asthere are 10 vertices in Figure 3, so, this is the last iteration.

So, by the Algorithm 2DIG, we get D = {4, 5, 8, 7, 9, 10}, that is, D = {4, 5, 7, 8, 9, 10}and L = {1, 6, 7, 9}. Therefore, |D| = cardinality of D = 6. In Figure 3, thick lines represent themembers of D.

4. Proof of Correctness and Time Complexity

Here, we will prove that D is a minimum 2-tuple dominating set.


Theorem 4.1. The set D is a minimal 2-tuple dominating set.

Proof. LetD = {d1, d2, . . . , dk} be the 2-tuple dominating set obtained by Algorithm 2DIG.Weare to prove that this D is minimal 2-tuple dominating set, that is, there does not exist anyD′ ⊂ D such that D′ is a 2-tuple dominating set.

Suppose, there exists a D′ ⊂ D such that D′ is a 2-tuple dominating set. Since D′ ⊂ D,there must exist at least one member of D, say di, such that di /∈ D′. Let the leading vertexcorresponding to di be k, then wk(f) = 2. Again, since D′ is a 2-tuple dominating set anddi /∈ D′, f(di) = 0 and wk(f) = 1 with respect to the 2-tuple dominating set D′. Therefore, kis dominated by only one vertex of D′, which is a contradiction of our assumption that D′ isa 2-tuple dominating set. Thus, D is minimal 2-tuple dominating set.

Theorem 4.2. The 2-tuple domination number of the given interval graph is the cardinality of the2-tuple dominating set D, that is, γ×2(G) = |D|.

Proof. Let L be the set of leading vertices corresponding to theminimal 2-tuple dominating setD of G. Suppose there exists another minimal 2-tuple dominating set D′ such that |D′| < |D|.

Without loss of generality, we assume that l1 is the leading vertex of order 2corresponding to the two vertices d1 and d2 of D. Then, l1 is adjacent to exactly two verticesd1, d2 (d1 < d2) of D. Also d2 = M0(l1), the highest numbered adjacent vertex to l1. So, theredoes not exist any vertex v > d2 in V such that l1 is adjacent to v. If d1, d2 /∈ D′, then thereexist at least two vertices, say, d′

1, d′2 ∈ D′ such that d′

1 < d′2 < d1 < d2, where each d′

1 andd′2 are adjacent to l1. If |D| = 2, then we have |D′| = 2. So, γ×2(G) = |D′| = |D|. For |D| > 2,

consider the following two cases.

Case 1. Let l2 be the leading vertex of order 1 corresponding to a vertex d3 ∈ D. Since l2 is oforder 1, either d3 = M0(l2) or d3 = M1(l2) (by Algorithm 2DIG) and l2 is adjacent to d2 butnot adjacent to d1. If l2 is adjacent to d1, then l2 is adjacent to three vertices d1, d2, and d3 ofD (not exactly two), a contradiction. Hence, l2 is not adjacent to the vertices d′

1 and d′2. As,

(d′1, l2) ∈ E or (d′

2, l2) ∈ E implies (d1, l2) ∈ E. Therefore, l2 ∈ V is not dominated by at leasttwo vertices ofD′ and hence there exist at least two vertices d′

3, d′4 ∈ D′, where each d′

3 and d′4

are adjacent to l2. Hence, |D′| ≥ |D|.

Case 2. Let l2 be the leading vertex of order 2 corresponding to the vertices d3, d4 ∈ D (d3 <d4). Then, l2 is not adjacent to any vertex higher than d4. Also, l2 is not adjacent to d′

1 and d′2

as l2 is not adjacent to d2. Therefore, if D′ is a 2-tuple dominating set, l2 must be dominatedby at least two vertices of D′, say d′

3 and d′4. Hence, |D′| ≥ |D|.

Thus, there does not exist anyD′ such that |D′| < |D|, that is,D is minimum and henceγ×2(G) = |D|.

Henceforth, D means the minimum 2-tuple dominating set and L is the set of leadingvertices corresponding to D.

Theorem 4.3. The 2-tuple dominating set of an interval graph can be computed sequentially inO(n2)time.

Proof. Let the processor take unit time to perform a single instruction. Step 1 of Algo-rithm 2DIG takes O(n) time. The algorithm consists of a loop from Step 2.1 to Step 2.3. This


loop carry over n times. Within this loop, we see that a loop occurs, which is terminated afterO(|N[i]|) times. It is clear that |N[i]| ≤ p ≤ n, p is the upper bound of |N[i]|, for fixed i. Inthe worst case, we assume the loop runs over n times. So the total time complexity of Step 2is O(n2). Hence, the overall time complexity of the Algorithm 2DIG is of O(n2).

5. Some Important Results Related to Minimum 2-Tuple Domination

In this section, we present some important results related to minimum 2-tuple dominationon interval graphs. For a given interval graph G, let a tree T(G) = (V, E′) be defined such thatE′ = {(u,H(u)) : u ∈ V, u/=n}, let n be the root of T(G). This tree is called the interval tree.The various properties of interval tree are available in [6, 10, 28].

The following lemma is true for every connected interval graph.

Lemma 5.1 (see [28]). For a connected interval graph, there exists a unique interval tree T(G).

For each vertex v of interval tree, level(v) is the distance of v from the vertex n in thetree. The height h of the tree T(G) is defined by

h = max{level(v) : v ∈ V }. (5.1)

We have found a result for the minimum 2-tuple dominating set D in terms of theheight h of interval tree T(G) stated as follows.

Lemma 5.2. Let T(G) be the interval tree of the interval graph G with height h, then

|D| ≥

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

2⌈h

3

⌉

, where h/= 3m for some m ∈ N,

2(h

3+ 1

)

, where h = 3m for some m ∈ N,

(5.2)

where N is the set of natural numbers.

Proof. From the definition of interval tree T(G), we know that the vertex 1 of V is at level h.By the property of interval tree T(G), we know that any vertex at level l is not adjacent witha vertex at level l − 2 and level (u) ≥ level (v), for every u < v, u, v ∈ V [8]. Therefore, it isclear that the neighbors of the vertex v of level l are either at level l or at level l − 1.

Let D = {v1, v2, . . . , vk} such that v1 < v2 < · · · < vk. As the vertices at level h are notadjacent with the vertices at level h − 2 or at level greater than h − 2, two vertices v1, v2 ofD must be taken from the level h or h − 1. For the least possible D, we assume that v2 is atlevel h − 1 and consequently v3 is either at level h − 1 or h − 2 or h − 3. If v2 is at level h, thenpossibility of having v3 is either at h or at h − 1 or at h − 2 which decreases the level fromearlier level and hence the number of vertices ofD may increase. So this last case is excludedfrom our result as the result demands the lower bound ofD. Also, in further cases, we neglectsuch cases for the same reason. Thus, we take v3 at level h − 3, v4 at level h − 4, v5 at levelh−6, v6 at level h−7, v7 at level h−9, and so on. That is, v2k+1 at level h−3k and v2k+2 at levelh − 3k − 1, for k = 0, 1, 2, 3, . . .. So for each k there are two vertices from the consecutive levelsh − 3k and h − 3k − 1.


Now, if h = 3m, for some m ∈ N, then h − 3k is the last level, that is, level 0 of T(G).So,

h − 3k = 0, this gives k =h

3. (5.3)

Thus, there are ((h/3) + 1) consecutive levels and hence the least value of |D| is 2((h/3) + 1).If h/= 3m, for somem ∈ N, then h− 3k is not at the last level of T(G). So one vertex is requiredat level h − 3k − 1 or h − 3k − 2. In this case, k = h/3� − 1. So there are 2 h/3� consecutivelevels and hence the least value of |D| is 2 h/3�.

Therefore,

|D| ≥

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

2⌈h

3

⌉

, where h/= 3m for some m ∈ N,

2(h

3+ 1

)

, where h = 3m for some m ∈ N.

(5.4)

Here, we are going to prove a result that removal of a vertex v from graph G, G − vand G have the same minimum 2-tuple dominating set D.

Lemma 5.3. Let v /∈ L ∪D and G′ = (V ′, E′), where V ′ = V − {v} and E′ = {(i, j) ∈ E : i ∈ V ′, j ∈V ′}. Then

(i) the minimum 2-tuple dominating set of G′ and G is D, if G′ is connected,

(ii) if G′ is disconnected with k components (blocks), say, G1, G2, . . . , Gk, then there must existminimum 2-tuple dominating sets D1, D2, . . . , Dk of G1, G2, . . . , Gk such that D = D1 ∪D2 ∪ · · · ∪Dk and Di’s are pairwise disjoint.

Proof. (i) SupposeG′ is connected. Since, v /∈ L∪D, that is, v ∈ V −L∪D. By Algorithm 2DIG,at kth iteration, say, either if or else-if condition is satisfied for a vertex k of V , then k ∈ L∪D,otherwise, k ∈ V −L∪D. In this case, v ∈ V −L∪D, at v-th iteration, else condition is satisfiedfor the vertex v which has no effect on L andD. Hence, if the vertex is being deleted from thegraph G, then the new induced subgraph G′ = G − {v} has the same 2-tuple dominating setD as G.

(ii) Let G′ be disconnected and G′ = G1 ∪ G2 ∪ · · · ∪ Gk, where G1 = (V1, E1), G2 =(V2, E2), . . . , Gk = (Vk, Ek). Let us decompose D into disjoint subsets D1, D2, . . . , Dk such thatD1 ⊆ V1, D2 ⊆ V2, . . . , Dk ⊆ Vk, where V = V1 ∪ V2 ∪ · · ·Vk ∪ {v}, that is, D = D1 ∪D2 ∪ · · · ∪Dk. As D is obtained by Algorithm 2DIG and v has no effect on D, then v has no effect onD1, D2, . . . , Dk, and they are also obtained by Algorithm 2DIG. Therefore, D1, D2, . . . , Dk areminimum 2-tuple dominating sets of interval graphs G1, G2, . . . , Gk, respectively.

The generalized form of the Lemma 5.3 is as follows.


Corollary 5.4. Let S = {v ∈ V : v /∈ L ∪D} and G′ = (V ′, E′), where V ′ = V − S and E′ = {(i, j) ∈E : i ∈ V ′, j ∈ V ′}. Then

(i) the minimum 2-tuple dominating set of G′ is also D, if G′ is connected,

(ii) if G′ is disconnected with k components (blocks), say, G1, G2, . . . , Gk, then there must existminimum 2-tuple dominating sets D1, D2, . . . , Dk of G1, G2, . . . , Gk such that D = D1 ∪D2 ∪ · · · ∪Dk and Di’s are pairwise disjoint.

Proof. (i) By Lemma 5.3, we have seen that the deletion of v /∈ L ∪ D does not change theminimum 2-tuple dominating set D. Let G1 be a graph obtained after the deletion of v1 ∈ S,so D is also the 2-tuple dominating set of G1. Again, v2 ∈ S is being deleted from the graphG1 and the graph G2 is obtained. It also has the same 2-tuple dominating set D as of G.Proceeding in this way, we obtain the graph G′ which has same 2-tuple dominating set as G.(ii) The proof of this case follows from (ii) of Lemma 5.3.

In Lemma 5.3 and Corollary 5.4, the graph G′ is a subgraph of the graph G induced byV ′ whose vertex set is V ′ and edge set is the set of those edges of G that have both ends inV ′. By keeping the statement of Corollary 5.4 in mind, we define new terms 2-tuple base graphand redundant vertex as follows.

Definition 5.5 (2-tuple base graph). Let a graphG′ = (V ′, E′) be induced subgraph of the graphG = (V, E), where V ′ ⊆ V , E′ ⊆ E. The graph G′ is called the 2-tuple base graph of the graphG if the vertex set V ′ = L ∪D and edge set E′ = {(i, j) ∈ E : i ∈ V ′, j ∈ V ′}, where L is the setof leading vertices corresponding to minimum 2-tuple dominating set D of G.

Note 1. If V ′ = V = L ∪ D, then the graph G′ is the same as G and hence the graph G is the2-tuple base graph of the graph itself.

The 2-tuple base graph and its interval representation of the graph of Figure 1 aregiven in Figures 4 and 5, respectively. Note that, in case of 2-tuple base graph, L∪D = V ′, butin case of original graph, in general, L ∪D/=V .

Definition 5.6 (Redundant vertex). Let G = (V, E) be a given interval graph. A vertex v ∈ V issaid to be redundant in G, if the minimum 2-tuple dominating setD of G − v is same as of G.

An important conclusion is drawn about 2-tuple base graph as follows.

Lemma 5.7. Every interval graph has a unique 2-tuple base graph.

Proof. Suppose there exist two distinct 2-tuple base graphs G′ = (V ′, E′) and G′′ = (V ′′, E′′) ofthe interval graph G. Then, either (i) V ′ /=V ′′ or (ii) E′ /=E′′. Let L be the set of leading verticescorresponding to the minimum 2-tuple dominating set ofG. SinceG′ is the 2-tuple base graphof G, L ∪D = V ′ and E′ = {(i, j) ∈ E : i ∈ V ′, j ∈ V ′}. Again, G′′ is the 2-tuple base graph of G.Then, L ∪D = V ′′ and E′′ = {(i, j) ∈ E : i ∈ V ′′, j ∈ V ′′}. So L ∪D = V ′ = V ′′ and hence E′ = E′′,which is a contradiction of our assumption. Therefore, G′ and G′′ are same.

Nowwe define a relation between two interval graphs and it is proved that the relationis an equivalence relation.


8

1 5

4 6

7 9

10

Figure 4: 2-tuple base graph of the graph G.

1

5

4 6

7

8 9

10

Figure 5: Interval representation of the 2-tuple base graph of the graph G.

Lemma 5.8. Let GI be the set of all interval graphs. Let a relation, denoted by ≈, and defined byG1 ≈ G2 ⇒ G1 and G2 have same 2-tuple base graph, for all G1, G2 ∈ GI . Then, the relation ≈ is anequivalence relation.

Proof. The relation is an equivalence relation since the following properties hold as well.

Reflexive

Since every graph has unique 2-tuple base graph, the same graph has the same 2-tuple basegraph. Therefore, G1 ≈ G1.

Symmetric

G1 ≈ G2 ⇒ G2 ≈ G1, for all G1, G2 ∈ GI . Since, G1 ≈ G2 means G1 and G2 have the same2-tuple base graph, then G2 and G1 have the same 2-tuple base graph, that is, G2 ≈ G1.

Transitive

If G1 ≈ G2 and G2 ≈ G3 holds for all G1, G2, G3 ∈ GI , then G1 ≈ G3 holds. Let us considerG1, G2 have the same 2-tuple base graph G′ and G2, G3 have the same 2-tuple base graphG′′. However we know every graph has unique 2-tuple base graph, G2 cannot have distinct2-tuple base graph G′ and G′′. So G2 and G3 have the 2-tuple base graph G′ same as each ofG1, G2. So G1 and G3 have the same 2-tuple base graph G′. So, the transitive property holdsfor each of GI .

Since all the properties of equivalence relation hold good in GI , then the relation ≈defined on GI is an equivalence relation.

Definition 5.9 (2-tuple equivalent). An interval graph G1 is said to be 2-tuple equivalent to aninterval graph G2 if G1 ≈ G2, that is, G1, G2 have the same 2-tuple base graph.


Definition 5.10 (2-tuple equivalent class). Let GI be a set of interval graphs. GI is said to be2-tuple equivalent class if any interval graph of GI is 2-tuple equivalent to each and everygraph of GI .

From the above notions, we have an important result about 2-tuple equivalent class.

Lemma 5.11. Equivalence relation, defined on interval graphs, makes the partition of the set ofinterval graphs into 2-tuple equivalent classes.

Proof. This result directly follows from the abstract algebra that every equivalence relation de-fined on a set makes the partition of the set into equivalent classes. Hence, the result follows.Particularly, the partitions can be found by the 2-tuple base graph. That is, we are trying tosay that among all interval graphs, for each 2-tuple base graph, there is a 2-tuple equivalentclass.

Next, we have an another important result regarding the leading vertex correspondingto 2-tuple dominating set D.

Lemma 5.12. For an interval graph G = (V, E),

|D| = |L| + n2, (5.5)

where n2 is the number of leading vertices of order 2.

Proof. Let n1 be the number of leading vertices of order 1 and let n2 be the number of leadingvertices of order 2. By definition of leading vertex, a leading vertex of order 1 correspondsto a single vertex of D and leading vertex of order 2 corresponds to two vertices of D. Sincethere are n1 leading vertices of order 1, then D has n1 vertices and also there are n2 leadingvertices of order 2, so D has n1 + 2n2 vertices. Therefore, |D| = n1 + 2n2. Now, n1 + n2 = |L|.Thus, |D| = |L| + n2.

6. Conclusion

In this paper, we have traced out to find the minimum 2-tuple dominating set on intervalgraphs. The algorithm we have designed in this paper can be generalized to find minimumk-tuple dominating set and k-tuple domination number. Further investigations can be doneby generalizing our Algorithm 2DIG to find k-tuple dominating set of an interval graph. Wethink it will reduce the next researcher’s labour.

Acknowledgment

The authors would like to thank the anonymous referees for valuable comments and alsoexpress appreciation of their constructive suggestions to improve the paper.


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